Arithmetic - Time & Work - Previous Year CAT/MBA Questions

Answer: 6

Explanation :

Ratio of time taken by Anu, Tanu and Manu is 5 : 8 : 10.
⇒ Ratio of efficiencies of Anu, Tanu and Manu = $\frac{1}{5}$ : $\frac{1}{8}$ : $\frac{1}{10}$ = 8 : 5 : 4
Let their efficiencies be 8x, 5x and 4x respectively per hour.

Total work done by them in 4 days = (8x + 5x + 4x) × 4 × 8 = 17x × 32 = 544x

Now, Anu and Tanu worked by 6 days working 6 hours 40 minute i.e., 6$\frac{2}{3}$ hours daily.
∴ Worked completed by Anu and Tanu = 13x × 6 × $\frac{20}{3}$ = 520x

⇒ Time taken by Manu to complete the remaining work = $\frac{544x-520x}{4x}$ = 6 hours.

Hence, 6.

Answer: Option B

Explanation :

Let the efficiency of each person be ‘e’ units/hour.

Given,
35% work is done by N men in 10 days working for 7 hours/day, while
65% work is done by N-10 men in 14 days working for 10 hours/day

⇒ $\frac{\mathrm{N}\times 10\times 7}{0.35}$= $\frac{(\mathrm{N}-10)\times 14\times 10}{0.65}$

⇒ $\frac{\mathrm{N}\times 7}{7}$= $\frac{(\mathrm{N}-10)\times 14}{13}$

⇒ N = $\frac{(\mathrm{N}-10)\times 14}{13}$

⇒ 13N = 14N – 140

⇒ N = 140

Hence, option (b).

Answer: 11

Explanation :

Bob takes 40 days to finish the work. Alex takes is twice as fast as Bob and hence will take half the time taken by Bob i.e., 20 days. Alex is also thrice as fast as Cole, hence Cole will take thrice the time taken by Alex, i.e., 60 days.

Time take by
Alex = 20 days
Bob = 40 days
Cole = 60 days

Let the total work to be done = 120 units.

∴ Efficiency of 
Alex = 6 units/day
Bob = 3 units/day
Cole = 2 units/day

​​​​​​​

Work done/cycle = 22 units.

∴ Work done in 5 cycles = 5 × 22 = 110 units

Work left after 5 cycles (15 days) = 120 – 110 = 10 units.

On 16th day Alex and Bob will together complete 9 units of work while remaining 1 unit of work will be completed by Bob and Cole on 17th day.

∴ Alex worked for 10 days in 5 complete cycles + on 16th day i.e., total 11 days.

Hence, 11.

Answer: Option D

Explanation :

Let the total work to be done = LCM (15, 12, 20) = 60 units.
Efficiency of Anu = 60/15 = 4 units/day
Efficiency of Vinu = 60/12 = 5 units/day
Efficiency of Manu = 60/20 = 3 units/day

Let us calculate the work done every 2 days.
Vinu works on both the days, hence work done by Vinu = 2 × 5 = 10 units
Anu works on alternate days, hence work done by Anu = 4 units
Manu works on alternate days, hence work done by Manu = 3 units

∴ Total work done in 2 days = 10 + 4 + 3 = 17 units.

Hence total work done in 6 days = 3 × 17 = 51 units.

On 7th day Vinu and Anu will work, hence work done = 5 + 4 = 9 units.

∴ Work done in total 7 days = 51 + 9 = 60 units [Work gets completed now]

∴ The number of days needed to complete the work is 7 days.

Hence, option (d).

Answer: 32

Explanation :

Let the total work to be done = LCM (12, 16, 24) = 48 units.

Let the efficiency of Amar, Akbar and Anthony be ‘x’, ‘y’ and ‘z’ units/month

⇒ x + y = 48/12 = 4   …(2)
⇒ y + z = 48/16 = 3   …(1)
⇒ z + x = 48/24 = 2   …(3)

Adding all these equations
⇒ x + y + z = 9/2 = 4.5   …(4)

Solving these four equations we get,
x = 1.5, y = 2.5 and z = 0.5

∴ The person who is neither slowest not fastest is Amar with efficiency of 1.5 units/month.

∴ Time required by Amar to complete the task alone = 48/1.5 = 32 months.

Hence, 32.

Answer: Option A

Explanation :

Let the total work to be done = LCM (60, 84) = 420 units.

Efficiency of Anil = 420/60 = 7 units/day
Efficiency of Bimal = 420/84 = 5 units/day

Work done by Anil in 10 + 14 days = 24 × 7 = 168 units

Work done by Bimal in 14 days = 14 × 5 = 70 units

∴ Work done by Charu = 420 – 70 - 168 = 182 units

Fraction of work done by Charu = 182/420 = 91/210

∴ Payment received by Charu = 91/210 × 21,000 = Rs. 9,100

Hence, option (a).

Answer: Option C

Explanation :

Let the filling and emptying capacity of A and B be ‘a’ and ‘b’ units/hour respectively.

Case 1: A is opened at 2 pm and B at 3 pm
Total work done till 10 pm = 8a - 7b

Case 2: A is opened at 2 pm and B at 4 pm
Total work done till 6 pm = 4a - 2b

Since work done is same in both cases, we have

8a – 7b = 4a – 2b

⇒ 4a = 5b

Now time taken by A alone to fill the tank = (Total work)/a = (4a – 2b)/a = (5b – 2b)/(5b/4) = 12/5 hours = 144 minutes.

Hence, option (c).

Answer: Option A

Explanation :

Let the efficiency of Rahul be ‘r’ units per hour and for Gautam be ‘g’ units per day.

Initially, Rahul works for 8 hours and Gautam works for 6 hours.
∴ Work done by them = 8r + 6g   …(1)

If both of them worked starting from 9 AM, they would have completed the work in 7.5 hours.
∴ Work done by them = 7.5r + 7.5g   …(2)

⇒ 8r + 6g = 7.5r + 7.5g
⇒ 0.5r = 1.5g
⇒ r = 3g

∴ Work to be done = 8r + 6g = 8r + 2r = 10r

∴ Time taken by Rahul to complete the work alone = 10r/r = 10 hours.

Hence, option (a).

Answer: 3000

Explanation :

Let the area to be painted = LCM(12, 16) = 48 units.

⇒ Efficiency of Anil = 48/12 = 4 units/day
and Efficiency of Barun = 48/16 = 3 units/day

Work done by Anil in 6 days = 6 × 4 = 24 units
Work done by Anil in 6 days = 6 × 3 = 18 units
∴ Remaining work done by Chandu = 48 – 24 – 18 = 6 units.

⇒ Payment received by Chandu = 24000/48 × 6 = Rs. 3,000

Alternately, 
Fraction of work done by Anil in 6 days = 6/12 = ½
Fraction of work done by Barun in 6 days = 6/16 = 3/8

⇒ Fraction of work done by Chandu in 6 days = 1 - ½ - 3/8 = 1/8

Since, Chandu completes 1/8^th of the work, he will receive 1/8^th of the payment.

∴ Chandu’s payment = 1/8 × 24,000 = Rs. 3,000

Hence, 3000.

Answer: 4

Explanation :

Let John’s efficiency be 1 unit/day.

∴ Jack’s efficiency is 2 units/day.

Jack and Jim’s efficiency is thrice John’s efficienc.

∴ 2 + e_Jim = 3 × 1 

⇒ e_Jim = 1

John takes three days more than that taken by three of them working together

Let the time taken by all three together is d.

∴ Total work to be done = 4 × d = 1 × (d + 3)

⇒ d = 1

∴ Jack finished the work in d + 3 = 4 days.

Since, Jack and Jim have same efficiency, Jim will also finish the work in 4 days.

Hence, 4.

Answer: 40

Explanation :

In 60 days 1.5 kms out of 6 kms is built.

∴ In 60 days 1.5/6 = 1/4^th work is done.

To complete the whole work, it would take 4 × 60 = 240 days, i.e., 180 more days.

Days remaining now is 200 – 60 = 140 days.

∴ To complete the remaining work in 140 days, contractor will have to hire more people. Assuming he hires x more men.

The work which 140 men would take 180 more days, now need to be completed by 140 + x men in 140 days.

∴ 180 × 140 = 140 × (140 + x)

⇒ x = 40

∴ Contractor hires 40 more people.

Hence, 40.

Answer: Option D

Explanation :

Let total work be the LCM of 12 and 9 = 36 units.

Let the efficiency of A and B be a and b respectively.

Work done per day when A and B are working together = 36/12 = 3 units.

∴ a + b = 3 ...(1)   

Work done per day when A is working at half efficieny and B is working at thrice efficiency = 36/9 = 4 units.

∴ (a/2) + 3b = 4 ...(2)

Solving (1) and (2), we get; a = 2. 

Time taken by A alone to complete the work = 36/2 = 18 days.

Hence, option (d).

Answer: 13

Explanation :

Let the work done by one man and one machine per day be x and y respectively.

Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job.

Since efficiency is inversely proportional to the time taken, so the efficiency of 3 men and 8 machines is twice that of 8 men and 3 machines.

∴ (3x + 8y) = 2(8x + 3y)

∴ 13x = 2y.

So, work done by 13 men in a day = work done by 2 machines in a day.

∴ If two machines can finish the job in 13 days, same work will be done by 13 men in 13 days.

Hence, 13.

Answer: Option A

Explanation :

Anil in one day can do 1/20^th of the work

Sunil in one day can do 1/40^th of the work

Anil starts the job and Sunil joins him after three days.

So, Anil would have done 3/20^th of the work by the time Sunil joins

After Sunil joins, they both would be doing 3/40^th of work everyday

Now, it is known that Bimal joins them after some days and finishes 10% of the work (i.e. 1/10^th of the work).

Now, Anil alone had done 3/20^th of the work in first 3 days and Bimal completes 1/10^th of the work

So, in total they would have done 3/20 + 1/10 = 1/4^th of the work.

Remaining work = 3/4^th, which would be done by Anil and Sunil together.

Anil and Sunil together comple 1/20^th + 1/40^th = 3/40^th work in 1 day.

Hence, time taken by them to comple 3/4^th of the work = 10 days.

Combining everything,

Total number of days = 3 + 10 = 13 days.

Hence, option (a).

Answer: 12

Explanation :

If John works the same number of regular and over-time hours say 'p'

The income would be 57p and 114p

Let's say that he works 'x' hours regular and 'y' hours overtime...

So, the income would be 57x and 114y

we are told that 114y is 15% of 57x

114y = 0.15 × 57x

y = 0.075x

we also know that x + y = 172

therefore, x + 0.075x = 1.075x = 172

x = 160

y = 172 - 160 = 12

Therefore, the number of hours he worked overtime is 12 hours.

Hence, 12.

Answer: 10

Explanation :

Let a filling pipe fills the tank at ‘a’ liters per hour and a draining pipe drains at ‘b’ liters per hour.

Work done by 6 filling and 5 emptying pipes in 6 hours = 6(6a - 5b)
Work done by 5 filling and 6 emptying pipes in 60 hours = 60(5a - 6b)

6(6a – 5b) = 60(5a – 6b)
∴ 6a – 5b = 50a – 60b
∴ 44a = 55b
∴ a = 5b/4

Let the tank gets filled completely in ‘m’ hours when one draining pipe and two filling pipes are on.
Work done by 2 filling and 1 emptying pipes in m hours = m(2a - b)

∴ m(2a – b) = 6(6a – 5b)

∴ $\mathrm{m}\left(2\times \frac{5}{4}\mathrm{b}-\mathrm{b}\right)$ = $6\left(6\times \frac{5}{4}\mathrm{b}-5\mathrm{b}\right)$

∴ $\mathrm{m}\left(\frac{6\mathrm{b}}{4}\right)$ = $6\left(\frac{10\mathrm{b}}{4}\right)$

Solving this, we get m = 10

Hence, 10.

Answer: Option B

Explanation :

Let work-done by a human and a robot in a day are ‘h’ and ‘r’ respectively.

Work done by 15 human and 5 robot in 30 days = 30(15h + 5r)
Work done by 5 human and 15 robot in 60 days = 60(5h + 15r)

∴ Total work = 30(15h + 5r) = 60(5h + 15r)
⇒ 15h + 5r = 10h + 30r
⇒ h = 5r

Let fifteen humans take ‘y’ days to finish the job.

Work done by 15 humans in y days = y × 15

∴ 30(15h + 5r) = y × 15h
⇒ 30(15h + h) = y × 15h
⇒ y = 32

Hence, option (b).

Answer: Option D

Explanation :

Ratio of time taken by A and B to complete a certain task = 4 : 5.
∴ Ratio of their efficiencies = 5 : 4.

Now, A completes 50% of the work alone and B completes 5% of the work alone. That means, A and B together complete 45% of the work together in 4 days.

Out of this 45%, work done by A and B individually will be in the ratio of their efficiencies.

∴ Work done by B in 4 days = 4/9^th of 45% = 20%. [Since the ratio of their efficiencies is 5 : 4]

⇒ B completes 20% (i.e., 1/5^th) of the work in 4 days.

Thus, B alone can finish the entire job in 20 days.

Hence, option (d).

Answer: 48

Explanation :

Suppose the inlet pipes of type A fill in water at the rate ‘a’ units per minute and the inlet pipes of type B fill in water at the rate ‘b’ units per minute.

Therefore we have the following

30(10a + 45b) = 60(8a + 18b)
∴ 300a + 1350b = 480a + 1080b
∴ 180a = 270b
∴ a = 1.5b

Total capacity of the tank
= 300a + 1350b = 300(1.5b) + 1350b
= 1800b

If 7 inlet pipes of type A and 27 inlet pipes of type B are opened, the volume of water filled in every minute

= 7a + 27b = 7(1.5b) + 27b = 37.5b

Therefore the number of minutes taken to fill the tank

$=\frac{\mathrm{1800b}}{37.\mathrm{5b}}=\mathrm{48\; minutes}$

Hence, 48.

Answer: Option C

Explanation :

Suppose ‘R’ and ‘G’ are the number of units of work completed by Ramesh and Ganesh everyday. Therefore the total work to be completed = 16(R + G) = 16R + 16G

For the first 7 days, both work at 100% efficiency and for the remaining 10 days, Ramesh works at 70% efficiency. Therefore the total work to be completed = 7(R + G) + 10(0.7R + G) = 14R + 17G

∴ 16R + 16G = 14R + 17G or G = 2R

Now if only Ganesh had worked after Ramesh fell sick:
Work to be done = 9(G + R) = 9(2R + R) = 27R
Efficiency of Ganesh = G = 2R

∴ Time taken to finish the remaining work = 27R/2R = 13.5 days.

Hence, option (c).

Answer: Option D

Explanation :

The time taken by pump A alone to fill the tank completely = t hours.

The time taken by pump B alone to fill the tank completely = ‘t - 2’ hours.

On Wednesday, pump A was used for 3 hour less and pump B was used for 2 hours to complete the remaining work.
⇒ Work done by A in 3 hours = Work done by B in 2 hours.
⇒ efficiency of A : efficiency of B = 2 : 3

Let efficiency of A be '2x' and that of B be '3x'.

Now, total work is done by B in (t - 2) hours and by A in t hours.
⇒ 3x × (t - 2) = 2x × t
⇒ 3(t - 2) = 2t
⇒ t = 6 hours

∴ The tank starts filling at 2 pm.

If both pumps start filling, their combined efficiency = 5x
Total work to be done = 3x(t - 2) = 12x

⇒ Time taken to fill the tank = 12x/5x = 2.4 hours = 2 hours 24 minutes.

Therefore on Thursday, the tank will be completely full at 4.24 pm.

Hence, option (d).

Answer: 15

Explanation :

Let the number of days required to complete the job be n.

1 person works on day 1, 2 on day 2, 3 on day 3, …. n on day n.

Using Unitary Method:
Each person has the same efficiency = $\frac{1}{120}$

Work done on
Day 1 = $1\times \frac{1}{120}$
Day 2 = $2\times \frac{1}{120}$
...
Day n = $\mathrm{n}\times \frac{1}{120}$

∴ Total Work = $1\times \frac{1}{120}$ + $2\times \frac{1}{120}$ + ... + $\mathrm{n}\times \frac{1}{120}$

This is also equal to 1.

∴ $\frac{1}{120}+\frac{2}{120}+\frac{3}{120}+..+\frac{n}{120}$ = 1

∑n = 120
⇒ $\frac{\mathrm{n}(\mathrm{n}+1)}{2}$ = 120
⇒ n = 15

Hence, 15.

Answer: Option A

Explanation :

In 1 hour, the inlet pipe will fill $\frac{1}{8}$ of the tank.

Let the time taken by the outlet pipe to empty the tank be ‘x’ hours. So in 1 hour,
The outlet pipe will empty $\frac{1}{x}$ of the tank in 1 hour.

Together they fill the tank in 10 hours.

⇒ $\frac{1}{8}$ - $\frac{1}{x}$ = $\frac{1}{10}$

∴ x = 40

⇒ Outlet pipe takes 40 hours to empty the full tank.
∴ Outlet pipe will take 20 hours to empty half the tank.

Hence, option (a).

Answer: Option A

Explanation :

In one day, Amal, Bimal and Kamal complete $\frac{1}{4}$ of the work.

In one day, Amal and Bimal complete  $\frac{1}{10}+\frac{1}{8}=\frac{4+5}{40}=\frac{9}{40}$ of the work.

So amount of work done by Kamal alone in one day is $\frac{1}{4}-\frac{9}{40}$ = $\frac{1}{40}$ of the work

∴ Fraction of work done by Kamal in 4 days = 4 × $\frac{1}{40}$ = $\frac{1}{10}$

⇒ Amount received by Kamal will be 1/10th of total amount = 1/10 × 1000 = Rs. 10

Hence, option (a).

Answer: Option C

Explanation :

The change in the amount of chemical in each tank after every minute is as follows:

A: −20 – 10 + 90 = 60

B: −100 + 110 + 20 = 30

C: −50 − 90 + 100 = −40

D: −110 + 10 + 50 = −50
 
Since tank D loses the maximum amount of chemical in a minute, it will be emptied first.

Let n minutes be the time taken by tank D to get empty.

∴ 1000 – 50n = 0

∴ n = 20 minutes

Hence, option (c).